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Is the meson color wave function symmetric of anti-symmetric? I know that in the case of the baryon it is anti-symmetric; however, I also know that baryons(which are made of three quarks) are fermions while mesons (which are composed of quark and antiquark) are bosons. Thanks.

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Mesons have valence quark content $q \bar q$ which does not comprise an identical pair of particles so indeed they are not susceptible to Fermi Dirac statistics but rather to Bose Einstein statistics. There is therefore no enforcement of antisymmetry upon the wavefunction description of the bound state. However, as an observable hadronic state, it must be a colour singlet and so for a quark $q_i$ and an antiquark $\bar q_j$, where $i$ and $j$ label the colour degree of freedoms, we must contract with the Kronecker delta $\delta_{ij}$ to project out the physically realisable configurations. That is to say, the colour component of the meson wavefunction will always be in the $SU(3)$ colour symmetric state $\frac{1}{\sqrt{3}}(r \bar r + b \bar b + g \bar g)$ as $$|\Psi \rangle = \frac{1}{\sqrt{6}}\sum_{i,j} \delta_{ij} q_i \bar q_j$$ entails. The relevant projector in colour space is the $\delta_{ij}$ which is manifestly symmetric.

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